Question

Ethanol, the alcohol used in automobile fuels, is produced by the fermentation of sugars present in plants. Corn is often used as the sugar source. The following equation represents the fermentation of glucose, the sugar in corn, by yeast to produce ethanol and carbon dioxide. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g) $$ Ethanol is combusted in an automobile engine according to the equation $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) $$ What would be the total volume of \(\mathrm{CO}_{2}\) gas formed at STP when \(3.00 \mathrm{~kg}\) of sugar is fermented and the ethanol is then combusted in an automobile engine?

Short answer

The total volume of \(\mathrm{CO}_2\) gas formed is approximately 2240.45 liters at STP.

Step-by-step solution

Balance the Equations

Verify the chemical equations given in the problem to ensure they are balanced. The fermentation equation: \( \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}(aq) \rightarrow 2 \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(aq) + 2 \mathrm{CO}_{2}(g) \), and the combustion equation: \( \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2}\mathrm{O}(g) \), are already balanced.

Calculate Moles of Glucose

Molar mass of glucose (\(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\)) is calculated as follows: \( 6\times12 + 12\times1 + 6\times16 = 180 \text{ g/mol}\). Convert 3.00 kg of glucose to grams (3000 g), then calculate moles: \( \text{Moles of glucose} = \frac{3000 \text{ g}}{180 \text{ g/mol}} = 16.67 \text{ moles}\).

Calculate CO2 from Fermentation

Each mole of glucose produces 2 moles of \(\mathrm{CO}_2\) during fermentation. Thus, \(16.67 \text{ moles of glucose} \times 2 = 33.34 \text{ moles of } \mathrm{CO}_2 \) from fermentation.

Calculate Ethanol from Fermentation

Each mole of glucose also produces 2 moles of \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\). Therefore, \(16.67 \text{ moles of glucose} \times 2 = 33.34 \text{ moles of } \mathrm{C}_2\mathrm{H}_5\mathrm{OH} \).

Calculate CO2 from Combustion

Each mole of \(\mathrm{C}_2\mathrm{H}_5\mathrm{OH}\) produces 2 moles of \(\mathrm{CO}_2\) in the combustion reaction. Thus, \(33.34 \text{ moles of } \mathrm{C}_2\mathrm{H}_5\mathrm{OH} \times 2 = 66.68 \text{ moles of } \mathrm{CO}_2 \).

Calculate Total CO2 Produced

Total \(\mathrm{CO}_2\) is the sum from fermentation and combustion: \(33.34 + 66.68 = 100.02 \text{ moles of } \mathrm{CO}_2 \).

Convert Moles of CO2 to Volume at STP

At STP, 1 mole of gas occupies 22.4 L. Thus, \(100.02 \text{ moles of } \mathrm{CO}_2 \times 22.4 \text{ L/mol} = 2240.448 \text{ L}\) of \(\mathrm{CO}_2\) gas.

Key concepts

Fermentation

Fermentation is a fascinating chemical process where sugars are converted into alcohol and carbon dioxide. This biological process occurs in the absence of oxygen and is catalyzed by yeast and other microorganisms. In our exercise, this is demonstrated with glucose (\(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\)), which is common in plant sources like corn.

• The fermentation reaction releases energy and involves breaking down glucose into two molecules of ethanol (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\)) and two molecules of carbon dioxide (\(\mathrm{CO}_{2}\)).
• Fermentation is an anaerobic, or oxygen-free, reaction that has been utilized in food and industrial processes for centuries. This process not only provides ethanol for fuel but also a variety of alcoholic beverages.
• It serves as a biological route to convert renewable plant materials into biofuels.

Combustion

Combustion is a high-energy reaction that involves the oxidation of substances, typically in the presence of oxygen, to release energy in the form of heat and light. In our context, ethanol, which is produced from the fermentation of glucose, is burned in an engine.

• The equation for the combustion of ethanol (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(l) + 3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + 3 \mathrm{H}_{2}\mathrm{O}(g)\)) shows that ethanol reacts with oxygen to form carbon dioxide and water.
• Combustion reactions are crucial in energy production, especially in engines where fuel is burned to harness power for movement.
• The perfect balance of fuel to oxygen ensures complete combustion, maximizing energy output while minimizing unburnt residues.
• Percentage yields and efficiencies in combustion help inform better industrial practices and reduction of pollution.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It enables chemists to predict the outcomes of reactions and calculate quantities needed.

• In the exercise, stoichiometry is used to determine how much \(\mathrm{CO}_{2}\) is produced when ethanol is combusted after fermentation of glucose.
• Moles are the standard unit of measurement in stoichiometry. We accurately calculated them from given masses and used the balanced chemical equations to understand the ratios of reactants to products.
• The molar mass of glucose helped convert a mass (3.00 kg) into moles, facilitating the prediction of product amounts using stoichiometry.
• Accurate balancing and calculation techniques provide exact results, ensuring efficient use of resources and minimizing waste in chemical processes.

Gas Laws

Gas laws are mathematical models that describe how gases behave, specifying their relations between volume, pressure, and temperature. They are crucial when dealing with gaseous products in chemical reactions.

• In the given problem, the Ideal Gas Law (\(PV = nRT\)) and the concept of Standard Temperature and Pressure (STP) are used to find the volume of \(\mathrm{CO}_{2}\) gas produced.
• When a gas is at STP, it occupies a fixed volume per mole, specifically 22.4 liters, which is employed to convert moles of gas to volume.
• Understanding gas laws allows for the accurate calculation of product volumes under different conditions, an essential aspect of industrial chemistry.
• Leveraging these laws effectively informs environmental management strategies, especially concerning emissions control.